In the real world (real solutions): $x^{2}=1$ $\implies$ $x=1,x=-1$ $x^{3}=1$ $\implies$ $x=1$ $x^{4}=1$ $\implies$ $x=1,x=-1$ $x^{5}=1$ $\implies$ $x=1$ $x^{6}=1$ $\implies$ $x=1,x=-1$
In the complex world: $x^{3}=1$ $\implies$ $x=1, x=-\frac{1}{2}+\frac{\sqrt{ 3 }}{2}, x=-\frac{1}{2}-\frac{\sqrt{ 3 }}{2}$ (note 3 solutions) $x^{4}=1$ $\implies$ $x=1,x=-1, x=i, x=-i$ (note 4 solutions)
Theorem: Including complex numbers, there are $n$ $nth$ roots of 1. They have modulus 1 and are spaced at angular intervals of $\frac{2\pi}{n}$.
More general:
Given a non-zero complex number $z$, the total numb of $nth$ roots of $z$ (including complex roots) is $n$.
The roots all have the same modulus, and are spaced at angular intervals of $\frac{2\pi}{2}$, i.e. they form the vertices of a regular $n$-gon
e.g.
Find all the $6th$ roots of $-8i$ given that $1+i$ is one of them.
Evaluate $(-3+4i)^{6}$
Recall that $r_{1}cis\theta_{1}\times r_{2}cis\theta_{2}=r_{1}r_{2}cis(\theta_{1}+\theta_{2})$
Then we get De Moivre's Theorem:
$(rcis\theta)^{n}=r^{n}cis(n\theta)$
e.g. Find the three cube roots of $z=4\sqrt{ 3 }+4i$, giving answers in exact polar form.
$$\begin{align} |z|=\sqrt{ 16\times{3} + 16 }&=8 , \tan \theta=\frac{4}{4\sqrt{ 3 }} \therefore \theta=\frac{\pi}{6} \\ \text{So }z&=8cis\frac{\pi}{6} \\ \therefore x^{3}&=8cis\frac{\pi}{6} \\ (x^{3})^{1/3}&=\left( 8cis\frac{\pi}{6} \right)^{1/3} \\ x&=8^{1/3}cis\left( \frac{1}{3}\times \frac{\pi}{6} \right) \\ &=2cis\left( \frac{\pi}{18} \right) \\ \text{But instead,} \\ x^{3}&=8cis\left( \frac{\pi}{6}+2k\pi \right) \\ &\dots \\ x&=2cis\left( \frac{\pi}{18}+\frac{2k\pi}{3} \right) \\ x&= 2cis\left( \frac{\pi}{18} \right),=2cis\left( \frac{13\pi}{18} \right),=2cis\left( -\frac{11\pi}{18} \right) \end{align} $$
